CAT 2020 Quant Questions — Slot 3

Think of this as: we need the first moment when both Anil and Sunil are back at the starting line simultaneously — then we just check how far Ravi has run by that moment.

Step 1 — Time each runner takes for one full lap.

The track is 3 km. A runner returns to the start each time they complete a full lap, so the lap-time is:

$\text{Lap-time} = \frac{\text{Track length}}{\text{Speed}}$

Step 2 — When are Anil and Sunil both at the start together?

Anil is at the start at times $0.2,\; 0.4,\; 0.6,\; 0.8,\; \ldots$ hours.

Sunil is at the start at times $0.3,\; 0.6,\; 0.9,\; 1.2,\; \ldots$ hours.

The first common time is the LCM of their lap-times:

$\text{LCM}(0.2,\; 0.3) = 0.6 \text{ hours} \;(= 36 \text{ minutes})$

💡 Teacher tip: "Meet again at the starting point" is always an LCM of lap-times problem. It has nothing to do with relative speeds or overtaking — it only asks: what is the smallest time that is a whole multiple of each runner's lap-time?

Step 3 — Verify this is the first coincidence.

At $t = 0.6$ hours:

• Anil has run $\frac{0.6}{0.2} = 3$ laps ✅ (integer — he is at the start)
• Sunil has run $\frac{0.6}{0.3} = 2$ laps ✅ (integer — he is at the start)

No smaller positive time gives whole laps for both (the next candidate below 0.6 would need to divide both 0.2 and 0.3, but their GCD is 0.1, and $0.1/0.3$ is not an integer), so 0.6 hours is indeed the first simultaneous return.

Step 4 — Distance Ravi has covered in 0.6 hours.

Ravi's speed is 8 km/hr, and he has been running for 0.6 hours:

$\text{Distance} = \text{Speed} \times \text{Time} = 8 \times 0.6 = 4.8 \text{ km}$

Note that Ravi does not need to be at the starting point — the question only asks how far he has run, which is simply speed × time elapsed.

Answer: $\boxed{4.8 \text{ km}}$ — Option A

💡 Takeaway: Whenever a question says "meet again at the starting point", reach for the LCM of individual lap-times. Contrast this with "meet again anywhere on the track", which uses relative speed. The phrasing tells you which tool to use.

Alternative approach — using lap counts directly.

Let $t$ be the meeting time. For both to be at the start, each must complete a whole number of laps:

• Anil's laps: $\frac{15t}{3} = 5t$ must be an integer.
• Sunil's laps: $\frac{10t}{3}$ must be an integer.

From $5t \in \mathbb{Z}$, we get $t = \frac{k}{5}$ for some positive integer $k$.

Substituting into Sunil's condition: $\frac{10 \cdot k/5}{3} = \frac{2k}{3}$ must be an integer, so $3 \mid k$. The smallest such $k$ is $3$, giving $t = \frac{3}{5} = 0.6$ hours.

Ravi's distance $= 8 \times 0.6 = 4.8$ km. Same result, arrived at through integer divisibility instead of an explicit LCM.

Think of this as a two-stage problem: first, use the "6 minutes late" clue to pin down the distance and the usual travel time; then, lay out the timeline of the special day and find the speed needed on the last leg.

Step 1 — Find the distance and usual time from the "6 minutes late" clue.

Let the distance to office be $d$ km.

• At her usual speed of $40$ km/hr, time taken $= \dfrac{d}{40}$ hours.
• At the slower speed of $35$ km/hr, time taken $= \dfrac{d}{35}$ hours.

The difference is $6$ minutes $= \dfrac{6}{60} = \dfrac{1}{10}$ hour:

$\frac{d}{35} - \frac{d}{40} = \frac{1}{10}$

$d\left(\frac{40 - 35}{35 \times 40}\right) = \frac{1}{10}$

$d \cdot \frac{5}{1400} = \frac{1}{10}$

$d = \frac{1}{10} \times \frac{1400}{5} = 28 \text{ km}$

So the distance is $28$ km, and the usual time is:

$\frac{28}{40} = 0.7 \text{ hr} = 42 \text{ min}$

💡 Teacher tip: The "6 minutes late" sentence is doing double duty — it pins down both the distance ($28$ km) and the usual time ($42$ min). Solve it once, then carry those two numbers through the rest of the problem.

Step 2 — Lay out the special day.

On this particular day, Vimla covers $\dfrac{2}{3}$ of the distance in $\dfrac{1}{3}$ of her usual time.

• Distance covered in the first leg $= \dfrac{2}{3} \times 28 = \dfrac{56}{3}$ km.
• Time taken for the first leg $= \dfrac{1}{3} \times 42 = 14$ min.

Step 3 — Account for the 8-minute stop.

She then stops for $8$ minutes. So the total time used so far:

$14 + 8 = 22 \text{ min}$

Since her total allowed time to reach on time is $42$ minutes, the time remaining is:

$42 - 22 = 20 \text{ min} = \frac{20}{60} = \frac{1}{3} \text{ hr}$

Step 4 — Find the required speed on the remaining distance.

The remaining distance is what is left after covering $\dfrac{2}{3}$ of the total:

$\text{Remaining distance} = \frac{1}{3} \times 28 = \frac{28}{3} \text{ km}$

She must cover $\dfrac{28}{3}$ km in $\dfrac{1}{3}$ hour. The required speed is:

$v = \frac{\text{distance}}{\text{time}} = \frac{28/3}{1/3} = \frac{28}{3} \times \frac{3}{1} = 28 \text{ km/hr}$

Quick check:

Everything adds up: total distance $= 28$ km, total time $= 42$ min — exactly on schedule.

Answer: $\boxed{28}$

💡 Closing insight: The answer comes out to $28$ km/hr — the same as the distance in km. That's not a coincidence: the remaining distance is $\frac{1}{3}$ of $28$ km, and the remaining time is also $\frac{1}{3}$ of an hour, so the $\frac{1}{3}$ factors cancel cleanly, leaving $28$.

Think of this as: we need to split the prime factors of $4^{2017}$ into two piles $a$ and $b$ (with $a \le b$). The question is really asking how many distinct ways we can choose the smaller factor.

Step 1 — Express $4^{2017}$ as a single prime power.

$4^{2017} = (2^2)^{2017} = 2^{4034}$

So we are looking for pairs $(a, b)$ of positive integers with $a \le b$ and $ab = 2^{4034}$.

Step 2 — Count the divisors of $2^{4034}$.

A number of the form $p^n$ has $n + 1$ divisors: $1, p, p^2, \ldots, p^n$.

So $2^{4034}$ has $4034 + 1 = 4035$ divisors.

Step 3 — Understand how divisors pair up.

For every divisor $a$ of $2^{4034}$, there is a complementary divisor $b = \frac{2^{4034}}{a}$ such that $ab = 2^{4034}$.

• If $a < b$, this gives one valid pair with $a \le b$.
• If $a > b$, it is the same pair counted from the other side — so we must not double-count.
• If $a = b$, this is a single, self-paired divisor.

💡 Teacher tip: Divisors always come in mirror pairs around $\sqrt{N}$. The only time a divisor pairs with itself is when $N$ is a perfect square — that is the square root.

Step 4 — Identify the middle (self-paired) divisor.

Since $4034$ is even, $2^{4034}$ is a perfect square:

$2^{4034} = (2^{2017})^2$

So $a = b = 2^{2017}$ is the one divisor where $a = b$.

Step 5 — Calculate the count.

• Total divisors: $4035$.
• One divisor ($2^{2017}$) pairs with itself → contributes 1 pair.
• Remaining divisors: $4035 - 1 = 4034$.
• These form $\frac{4034}{2} = 2017$ pairs with $a < b$.

Total pairs with $a \le b$:

$2017 + 1 = 2018$

Verification with a smaller example.

Let's check the logic with $4^2 = 16 = 2^4$:

That gives $3$ pairs. Using our formula: $\frac{5 + 1}{2} = 3$. ✓ The pattern holds.

Alternative approach — direct formula.

For any $N$, the number of pairs $(a, b)$ with $a \le b$ and $ab = N$ equals $\left\lceil \frac{d(N)}{2} \right\rceil$, where $d(N)$ is the divisor count.

When $N$ is a perfect square, $d(N)$ is odd, so this becomes $\frac{d(N) + 1}{2}$.

Here, $d(2^{4034}) = 4035$ (odd, since $N$ is a perfect square), so:

$\frac{4035 + 1}{2} = 2018$

Answer: $\boxed{2018}$

💡 Why this is elegant: Notice the answer is $2017 + 1$. For $4^n = 2^{2n}$, the divisor count is $2n + 1$, and the number of pairs with $a \le b$ is $\frac{2n+1+1}{2} = n + 1$. So $4^{2017}$ gives exactly $2018$ pairs — the exponent plus one.

Think of this as: one train gets a 90-minute head start, so by the time the second train begins, the gap between them has already shrunk. We need to find the exact clock time when that remaining gap closes.

Step 1 — Account for the head start.

The train from A starts at 9:00 am at 40 km/hr. The train from B does not start until 10:30 am. So the first train travels alone for: $10{:}30 - 9{:}00 = 1.5 \text{ hours}.$

Distance covered by the train from A in those 1.5 hours: $40 \times 1.5 = 60 \text{ km}.$

Since the stations are 90 km apart, the remaining gap between the two trains at 10:30 am is: $90 - 60 = 30 \text{ km}.$

💡 Teacher tip: Whenever two objects start at different times, freeze the early starter at the later start time. The problem then collapses into a clean closing-speed question on whatever gap remains.

Step 2 — Closing speed after 10:30 am.

From 10:30 am onward, both trains move toward each other. Their relative (closing) speed is the sum of their speeds: $40 + 20 = 60 \text{ km/hr}.$

Step 3 — Time to close the remaining 30 km. $\text{Time} = \frac{\text{Distance}}{\text{Closing speed}} = \frac{30}{60} = 0.5 \text{ hours} = 30 \text{ minutes}.$

Step 4 — Clock time of meeting. $10{:}30 \text{ am} + 30 \text{ min} = 11{:}00 \text{ am}.$

Quick verification. By 11:00 am the train from A has run 2 hours: $40 \times 2 = 80$ km from A. The train from B has run 30 minutes: $20 \times 0.5 = 10$ km from B. Together: $80 + 10 = 90$ km — the full distance. ✓

Answer: $\boxed{11{:}00 \text{ am}}$ — Option C

Alternative approach — single equation. Let the trains meet $t$ hours after 9:00 am. Then the train from A has travelled $40t$ km, and the train from B (which started 1.5 h later) has travelled $20(t - 1.5)$ km. Their distances must add to 90: $40t + 20(t - 1.5) = 90 \Rightarrow 60t - 30 = 90 \Rightarrow t = 2.$ So the meeting time is $9{:}00 + 2$ hours $= 11{:}00$ am. Same answer, slightly more algebra.

💡 Takeaway: The "freeze the early starter" method is faster under exam pressure because it avoids setting up an equation with a shifted time variable — but the equation method is a reliable cross-check.

Think of this as: we have 120 integers, and we want to remove every number that's divisible by 2, 5, or 7. Whatever's left is our answer. So we first count what gets removed.

Step 1 — Set up the complement

It's easier to count the numbers that are divisible by at least one of 2, 5, 7, then subtract that from 120.

Let:

• $A$ = multiples of 2 in $\{1, \ldots, 120\}$
• $B$ = multiples of 5
• $C$ = multiples of 7

We want $120 - |A \cup B \cup C|$.

Step 2 — Count each set and its overlaps

The number of multiples of $d$ up to 120 is $\lfloor 120/d \rfloor$.

A number divisible by both 2 and 5 is divisible by $\text{lcm}(2,5)=10$; similarly for the other pairs and the triple.

💡 Teacher tip: When the divisors are all prime (or coprime in pairs), the lcm of any subset is just their product. That's why $\text{lcm}(2,5)=10$, $\text{lcm}(2,7)=14$, $\text{lcm}(5,7)=35$, and $\text{lcm}(2,5,7)=70$.

Step 3 — Apply inclusion–exclusion

The size of the union is:

$|A \cup B \cup C| = |A|+|B|+|C| - |A\cap B| - |A\cap C| - |B\cap C| + |A\cap B\cap C|$

Plugging in:

$|A \cup B \cup C| = 60 + 24 + 17 - 12 - 8 - 3 + 1 = 79$

So 79 numbers in $\{1, \ldots, 120\}$ are divisible by at least one of 2, 5, 7.

Step 4 — Subtract from the total

$120 - 79 = \boxed{41}$

Answer: 41 (Option B)

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Alternative approach — using periodicity (Euler's totient)

A number is divisible by none of 2, 5, 7 exactly when it is coprime to $70 = 2\times5\times7$. Coprimality to 70 repeats every 70 numbers.

In $1$ to $70$: the count of numbers coprime to 70 is Euler's totient:

$\varphi(70) = 70\left(1-\tfrac{1}{2}\right)\left(1-\tfrac{1}{5}\right)\left(1-\tfrac{1}{7}\right) = 70 \cdot \tfrac{1}{2} \cdot \tfrac{4}{5} \cdot \tfrac{6}{7} = 24$

In $71$ to $120$ (which is 50 more numbers): these have the same residues mod 70 as $1$ to $50$. So we count numbers in $\{1,\ldots,50\}$ coprime to 70.

Using inclusion–exclusion on 50:

$\lfloor 50/2\rfloor + \lfloor 50/5\rfloor + \lfloor 50/7\rfloor - \lfloor 50/10\rfloor - \lfloor 50/14\rfloor - \lfloor 50/35\rfloor + \lfloor 50/70\rfloor$ $= 25 + 10 + 7 - 5 - 3 - 1 + 0 = 33$

Numbers coprime to 70 in $\{1,\ldots,50\}$: $50 - 33 = 17$.

Total: $24 + 17 = 41$ ✓ — same answer, good cross-check.

💡 Closing insight: The inclusion–exclusion method is faster here since we only have three primes and the floor divisions are quick. The totient/periodicity method shines when the range is very large compared to the product of primes, because you can count full cycles and handle just the remainder.

Think of this as: we know one diagonal of a parallelogram, and we're told the other two vertices sit on a fixed line. The bridge between the two diagonals is the centre of the parallelogram.

Step 1 — Translate "opposite vertices of a parallelogram" into a usable fact.

The single most important property here: the diagonals of a parallelogram bisect each other. So the midpoint of the diagonal joining the two unknown vertices equals the midpoint of the diagonal joining the two given vertices.

Let the given opposite vertices be $A=(2,1)$ and $C=(-3,-4)$. The centre of the parallelogram is the midpoint of $AC$:

$$M=\left(\frac{2+(-3)}{2},\,\frac{1+(-4)}{2}\right)=\left(-\tfrac12,\,-\tfrac32\right).$$

💡 Teacher tip: Whenever a problem says "opposite vertices of a parallelogram," your reflex should be: find the midpoint — that's the centre, and the other diagonal passes through it too.

Step 2 — Set up the unknown vertices.

Let the other two vertices be $B=(x_1,y_1)$ and $D=(x_2,y_2)$. Both lie on the line $x+9y+c=0$, so each satisfies the line equation:

$$x_1+9y_1+c=0,\qquad x_2+9y_2+c=0.$$

Step 3 — Use the midpoint condition on $B$ and $D$.

Since $M$ is also the midpoint of $BD$:

$$\frac{x_1+x_2}{2}=-\tfrac12 \;\Rightarrow\; x_1+x_2=-1, \qquad \frac{y_1+y_2}{2}=-\tfrac32 \;\Rightarrow\; y_1+y_2=-3.$$

Step 4 — Add the two line equations.

This is the clean move: we don't need to find $B$ and $D$ individually — we only need their sums, which we already have.

$$(x_1+9y_1+c)+(x_2+9y_2+c)=0$$ $$(x_1+x_2)+9(y_1+y_2)+2c=0.$$

Substitute $x_1+x_2=-1$ and $y_1+y_2=-3$:

$$(-1)+9(-3)+2c=0\;\Rightarrow\;-1-27+2c=0\;\Rightarrow\;-28+2c=0.$$

Step 5 — Solve for $c$.

$$2c=28\;\Rightarrow\;c=14.$$

Answer: $\boxed{14}$

💡 Takeaway: "Opposite vertices of a parallelogram" is a standing instruction to use midpoint of diagonals = centre. Once you have the centre, adding the two line equations is enough — you never even need to find the individual vertices $B$ and $D$.

Think of this as: two lines in the plane meet at exactly one point unless they are parallel or identical. So we need the condition under which these two lines are not parallel — which, for a general pair of linear equations, comes down to checking a single determinant.

What we want: the condition on $k$ so that the system $kx + y = 3, \qquad 4x + ky = 4$ has exactly one solution.

Step 1 — Recall the unique-solution rule.

For a general pair of linear equations $a_1 x + b_1 y = c_1, \qquad a_2 x + b_2 y = c_2,$ the system has a unique solution exactly when the determinant of the coefficient matrix is non-zero: $\Delta = a_1 b_2 - a_2 b_1 \neq 0.$ If $\Delta = 0$, the two lines are either parallel (no solution) or identical (infinitely many solutions) — in either case, not a unique solution.

Step 2 — Identify the coefficients and compute $\Delta$.

Here $a_1 = k$, $b_1 = 1$, $a_2 = 4$, $b_2 = k$, so $\Delta = (k)(k) - (4)(1) = k^2 - 4.$

Step 3 — Require $\Delta \neq 0$.

$k^2 - 4 \neq 0 \;\Longrightarrow\; k^2 \neq 4 \;\Longrightarrow\; k \neq 2 \text{ and } k \neq -2.$

💡 Teacher tip: The determinant vanishes precisely when the two rows of the coefficient matrix are proportional. Here that happens when $\frac{k}{4} = \frac{1}{k}$, i.e. $k^2 = 4$ — the same condition we just derived.

Step 4 — Write the condition compactly.

Saying "$k \neq 2$ and $k \neq -2$" is the same as saying $k$ is neither $+2$ nor $-2$, which is exactly $|k| \neq 2.$

Quick check of the excluded values:

So at $k = \pm 2$ the system fails to have a unique solution; for every other $k$ it succeeds.

Answer: $\boxed{|k| \neq 2}$ — option A.

💡 Takeaway: for $a_1 x + b_1 y = c_1,\; a_2 x + b_2 y = c_2$, the unique-solution test is always $a_1 b_2 \neq a_2 b_1$ — i.e. the rows aren't proportional. Here that becomes $k^2 \neq 4$, or equivalently $|k| \neq 2$.

What we want: $f(10) - f(-10)$, given the functional equation $f(x+y) = f(x)f(y)$ and the single value $f(5) = 4$.

Think of this as: the rule $f(x+y) = f(x)f(y)$ tells us how the function's values combine under addition. We just need to leverage that rule to reach $f(10)$ and $f(-10)$ from the one value we know.

Step 1 — Find $f(0)$. Set $x = 0$ and $y = 0$ in the functional equation: $f(0) = f(0+0) = f(0) \cdot f(0) = f(0)^2.$ So $f(0)^2 - f(0) = 0$, giving $f(0)\bigl(f(0) - 1\bigr) = 0$. Thus $f(0) = 0$ or $f(0) = 1$.

But if $f(0) = 0$, then $f(x) = f(x+0) = f(x)\cdot f(0) = 0$ for every $x$, which contradicts $f(5) = 4$. Therefore: $f(0) = 1.$

Step 2 — Compute $f(10)$. Since $10 = 5 + 5$: $f(10) = f(5+5) = f(5) \cdot f(5) = 4 \times 4 = 16.$

Step 3 — Find $f(-5)$, then $f(-10)$. Since $0 = 5 + (-5)$: $1 = f(0) = f\bigl(5 + (-5)\bigr) = f(5) \cdot f(-5) = 4 \cdot f(-5).$ So $f(-5) = \dfrac{1}{4}$.

Now since $-10 = (-5) + (-5)$: $f(-10) = f\bigl((-5)+(-5)\bigr) = f(-5) \cdot f(-5) = \frac{1}{4} \times \frac{1}{4} = \frac{1}{16}.$

💡 Teacher tip: The key insight is that $f(-x) = \frac{1}{f(x)}$ whenever $f(0) = 1$. This follows for free from the functional equation — you never need to find the base of the exponential explicitly.

Step 4 — Subtract. $f(10) - f(-10) = 16 - \frac{1}{16} = \frac{256}{16} - \frac{1}{16} = \frac{255}{16} = 15.9375.$

Verification check:

Answer: $\boxed{15.9375}$

Alternative approach — Find the base explicitly. The equation $f(x+y) = f(x)f(y)$ is the defining property of exponential functions: $f(x) = a^x$ for some base $a > 0$. From $f(5) = 4$: $a^5 = 4 \implies a = 4^{1/5}.$ Then: $f(10) = a^{10} = (a^5)^2 = 4^2 = 16, \qquad f(-10) = a^{-10} = \frac{1}{a^{10}} = \frac{1}{16}.$ So $f(10) - f(-10) = 16 - \frac{1}{16} = 15.9375$ — the same result. The first method is faster under exam conditions because it never requires identifying the base; it uses the functional equation's properties directly.

💡 Why this is elegant: We never needed to know what $f(x)$ looks like for arbitrary $x$. The functional equation alone — plus the single value $f(5) = 4$ — pinned down everything we needed through multiplication and reciprocals.

Think of this as: three fractions are all squeezed into the same narrow band $(0.2, 0.5)$, and we need to find the natural numbers $m, n$ (with $n$ even) that make all three true simultaneously.

What we want: the value of $m - 2n$.

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Step 1 — Pin down $n$ first (it is the most constrained variable).

The third fraction gives us $n$ directly: $0.2 < \frac{n}{11} < 0.5$

Multiply through by $11$: $2.2 < n < 5.5$

Since $n$ is a natural number: $n \in \{3, 4, 5\}$.

We are told $n$ is even, so $n = 4$.

💡 Teacher tip: When several variables share chained bounds, always solve for the most constrained one first. Here $n$ is locked by two conditions — the inequality and evenness — so it collapses to a single value immediately.

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Step 2 — Narrow down $m$ using the first fraction.

From $0.2 < \frac{m}{20} < 0.5$, multiply through by $20$: $4 < m < 10$

Since $m$ is a natural number: $m \in \{5, 6, 7, 8, 9\}$.

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Step 3 — Use the middle fraction $\frac{n}{m} = \frac{4}{m}$ to pin $m$ exactly.

We need $0.2 < \frac{4}{m} < 0.5$. Split into two inequalities:

• Upper bound: $\frac{4}{m} < 0.5 \Rightarrow 4 < 0.5m \Rightarrow m > 8$.
• Lower bound: $\frac{4}{m} > 0.2 \Rightarrow 4 > 0.2m \Rightarrow m < 20$.

From Step 2, $m \in \{5,6,7,8,9\}$. The upper bound $m > 8$ leaves only $m = 9$. The lower bound $m < 20$ is automatically satisfied.

So $m = 9,\; n = 4$.

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Step 4 — Verify all three conditions with $(m, n) = (9, 4)$.

Also, $n = 4$ is even ✅ and both $m, n$ are natural numbers ✅.

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Step 5 — Compute the answer. $m - 2n = 9 - 2(4) = 9 - 8 = 1$

Answer: $\boxed{1}$

💡 Closing insight: The key was the order of attack. By solving for $n$ first (the variable with the extra evenness constraint), the entire problem reduced to a simple range check on $m$. Whenever a problem gives you a ratio bound plus a parity or divisibility condition, anchor on that variable first — the extra filter does the heavy lifting for you.

Think of this as: three exponential expressions are pinned to the same value, and we need to extract a clean number from a combination of their exponents. The trick is to turn those exponents into logarithms so they start behaving like ordinary numbers.

Step 1 — Introduce a common value and convert to logarithms.

Let the common value be $k$: $14^a = 36^b = 84^c = k.$

Rewriting each equality in logarithmic form: $a = \log_{14} k, \quad b = \log_{36} k, \quad c = \log_{84} k.$

Now invert each one using the reciprocal property $\dfrac{1}{\log_p q} = \log_q p$: $\frac{1}{a} = \log_k 14, \quad \frac{1}{b} = \log_k 36, \quad \frac{1}{c} = \log_k 84.$

💡 Teacher tip: Whenever you see $x^a = y^b = z^c$, the move $\frac{1}{\text{exponent}} = \log_k(\text{base})$ is almost always the unlock. It converts awkward exponents into logarithms that share the same base $k$, so you can add, subtract, and combine them freely.

Step 2 — Simplify the bracket $\left(\dfrac{1}{c} - \dfrac{1}{a}\right)$.

Substitute the logarithmic forms: $\frac{1}{c} - \frac{1}{a} = \log_k 84 - \log_k 14.$

Using the quotient rule $\log_k x - \log_k y = \log_k\!\left(\dfrac{x}{y}\right)$: $\log_k 84 - \log_k 14 = \log_k\!\left(\frac{84}{14}\right) = \log_k 6.$

So the bracket collapses to a single logarithm: $\dfrac{1}{c} - \dfrac{1}{a} = \log_k 6$.

Step 3 — Bring in the factor $6b$.

We have $b = \log_{36} k$, so: $6b\!\left(\frac{1}{c} - \frac{1}{a}\right) = 6 \cdot (\log_{36} k) \cdot (\log_k 6).$

Now apply the chain rule for logarithms, $\log_p q \cdot \log_q r = \log_p r$ — the intermediate base $k$ cancels: $(\log_{36} k)(\log_k 6) = \log_{36} 6.$

Therefore: $6b\!\left(\frac{1}{c} - \frac{1}{a}\right) = 6 \log_{36} 6.$

Step 4 — Evaluate $\log_{36} 6$.

Since $36 = 6^2$, we have $6 = 36^{1/2}$, so: $\log_{36} 6 = \log_{36}\!\left(36^{1/2}\right) = \frac{1}{2}.$

Step 5 — Final calculation. $6 \times \frac{1}{2} = 3.$

Answer: $\boxed{3}$

💡 Why this works so cleanly: the numbers were not random — $84 = 14 \times 6$ is what made the bracket reduce to $\log_k 6$, and $36 = 6^2$ is what made the final logarithm equal $\frac{1}{2}$. The factor $6$ in front was chosen precisely to cancel that $\frac{1}{2}$.

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Alternative approach — pure exponent algebra (no logarithms).

From $14^a = 36^b = 84^c = k$, raise both sides of each equality to the reciprocal power to isolate the bases: $14 = k^{1/a}, \quad 36 = k^{1/b}, \quad 84 = k^{1/c}.$

Now use the number relationship $84 = 14 \times 6$: $k^{1/c} = k^{1/a} \cdot 6 \;\Longrightarrow\; 6 = k^{\,1/c \,-\, 1/a}.$

Also use $36 = 6^2$: $k^{1/b} = 6^2 = \left(k^{\,1/c \,-\, 1/a}\right)^2 = k^{\,2(1/c \,-\, 1/a)}.$

Equating exponents (since $k \neq 1$): $\frac{1}{b} = 2\!\left(\frac{1}{c} - \frac{1}{a}\right) \;\Longrightarrow\; \frac{1}{c} - \frac{1}{a} = \frac{1}{2b}.$

Therefore: $6b\!\left(\frac{1}{c} - \frac{1}{a}\right) = 6b \cdot \frac{1}{2b} = 3.$

This route is faster once you spot $84 = 14 \times 6$ and $36 = 6^2$; the logarithm method is more mechanical and safer under exam pressure.

Think of this as: we know one base, the height, and one angle — that's enough to find the other base, and then the area follows directly.

What we're given:

• $AB \parallel DC$, $BC \perp DC$
• $\angle BAD = 45°$, $DC = 5$ cm, $BC = 4$ cm

Since $BC \perp DC$ and $AB \parallel DC$, the side $BC$ is the height of the trapezium: $h = 4$.

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Step 1 — Drop a perpendicular from $A$ to line $DC$

Let $P$ be the foot of the perpendicular from $A$ to the line through $D$ and $C$ (extended if necessary).

Since $AB \parallel DC$ and $AP \perp DC$, we have $AP \perp AB$, so: $\angle BAP = 90°.$

Both $AP$ and $BC$ are perpendicular to $DC$, and $AB \parallel DC$, so they are equal in length: $AP = BC = 4.$

Step 2 — Use $\angle BAD = 45°$ to find $\angle PAD$

The angle $\angle BAD = 45°$ sits between sides $AB$ and $AD$. Since $AP \perp AB$ (from Step 1), the ray $AD$ lies between $AB$ and $AP$, giving: $\angle PAD = \angle BAP - \angle BAD = 90° - 45° = 45°.$

Step 3 — Solve the right triangle $APD$

In $\triangle APD$: $\angle APD = 90°$, $\angle PAD = 45°$, so the third angle is also $45°$ — it's an isosceles right triangle.

Therefore the two legs are equal: $PD = AP = 4.$

💡 Teacher tip: A $45°$ angle in a right triangle immediately forces an isosceles right triangle — the two legs are always equal. This is the key that unlocks the missing base length.

Step 4 — Locate $P$ and find $AB$

Since $\angle BAD = 45°$ is acute, the slanted side $AD$ must lean outward — meaning $P$ falls to the left of $D$, outside segment $DC$.

If $P$ were between $D$ and $C$, then $\angle BAD$ would be $90° + 45° = 135°$ (obtuse), contradicting the given $45°$.

So the horizontal layout is: $P$ — $D$ — $C$, with $PD = 4$ and $DC = 5$.

Since $A$ is directly above $P$ and $B$ is directly above $C$ (both $AB$ and $PC$ are horizontal), the top base is: $AB = PC = PD + DC = 4 + 5 = 9.$

Step 5 — Compute the area

$\text{Area} = \frac{1}{2}(AB + DC) \times h = \frac{1}{2}(9 + 5) \times 4 = \frac{1}{2} \times 14 \times 4 = 28.$

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Verification check:

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Answer: $\boxed{28}$

💡 Why this works: The $45°$ angle creates an isosceles right triangle whose leg equals the height. That leg extends the base by exactly $4$ cm beyond $D$, turning the unknown base $AB$ into a simple sum: $4 + 5 = 9$. The area then falls out in one line.

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Alternative approach — Coordinates. Place $D=(0,0)$, $C=(5,0)$. Since $BC\perp DC$ and $BC=4$, set $B=(5,4)$. Since $AB\parallel DC$ (horizontal), $A=(a,4)$. At vertex $A$, vectors $\vec{AB}=(5-a,0)$ and $\vec{AD}=(-a,-4)$. Setting $\tan 45°=1$: $1=\frac{|\vec{AB}\times\vec{AD}|}{\vec{AB}\cdot\vec{AD}}=\frac{4(5-a)}{-a(5-a)}=\frac{4}{-a}\;\Rightarrow\; a=-4.$ So $A=(-4,4)$, giving $AB=9$ and the same area $\frac{1}{2}(9+5)\times4=28$. The geometric method is faster under exam time; the coordinate method is a reliable cross-check when the configuration is ambiguous.

Think of this as: we know the overall average and the last two scores, and we know the average of the first $n$ scores — that's enough to pin down $n$. Once $n$ is fixed, we have a fixed total to split among the first $n$ innings, and we want the smallest possible lowest score.

Step 1 — Use the two averages to find $n$.

Across all $n+2$ innings the average is $29$, so the total runs scored are $29(n+2).$

The last two innings contribute $38 + 15 = 53$ runs. So the first $n$ innings contribute $29(n+2) - 53 = 29n + 58 - 53 = 29n + 5.$

But we are also told the first $n$ innings have an average of $30$, so their total is $30n$. Equating the two expressions for the same total: $29n + 5 = 30n \;\Rightarrow\; n = 5.$

So the batsman played $5 + 2 = 7$ innings in all, and the first $5$ innings sum to $30 \times 5 = 150$ runs.

Step 2 — Set up the minimisation.

We now have five positive integer scores (he got out every time, so each is a genuine completed score) that

• sum to $150$, and
• are each strictly less than $38$.

We want the smallest possible value of the lowest score $x$.

💡 Teacher tip: In any "minimise the minimum" problem, the winning move is to maximise everything else. Push the other four scores to their ceiling; the leftover is the smallest the minimum can be.

Step 3 — Maximise the other four innings.

Each of the first five innings must be $< 38$, i.e. at most $37$. To make the lowest score as small as possible, set the other four innings to their maximum allowed value, $37$ each: $x = 150 - 4 \times 37 = 150 - 148 = 2.$

Step 4 — Verify the boundary.

The set $\{2, 37, 37, 37, 37\}$ sums to $150$, every entry is $< 38$, and the lowest is $2$. Trying to push the lowest below $2$ would force at least one of the other four innings above $37$, breaking the $<38$ rule.

Answer: $\boxed{2}$ (Option C)

💡 Takeaway: "Minimise the minimum" problems always flip into "maximise everything else" — push the other variables to their ceiling and the minimum falls out by subtraction. The verification table is the quick way to confirm the boundary is tight.